Scalar and vector potentials
Problem:
(a) Write down Maxwell's equations in free space and in the presence of
the current density j(r,t) and charge density ρ(r,t). Introduce the electromagnetic potentials and derive
the differential equation that they satisfy.
(b) State the Lorentz condition and show the simplification found thereby.
(c) Define a gauge transformation and describe its effect on the
electromagnetic fields.
Solution:
 Concepts:
Maxwell's equations, scalar and vector potential, the Lorentz condition,
gauge transformations
 Reasoning:
Starting from Maxwell's equations, we are asked to introduce the
electromagnetic potentials and derive the differential equation that they
satisfy.
 Details of the calculation:
(a) Maxwell's equations in SI units are
∇∙E = ρ/ε_{0},
∇×E = ∂B/∂t, ∇∙B = 0,
∇×B = μ_{0}j + (1/c^{2})∂E/∂t.
We find
∇∙B = 0 >
B
=
∇×A,
∇×(E + ∂A/∂t) = 0 > >
E + ∂A/∂t
= ∇Φ.
Therefore
∇^{2}Φ + ∂∇∙A/∂t
= ρ/ε_{0}, and ∇×(∇×A)
 (1/c^{2})∂(∇Φ  ∂A/∂t)/∂t = μ_{0}j.
are the differential equations satisfied by the potentials.
Using
∇×(∇×A)
= ∇(∇∙A)
 ∇^{2}A
we write
∇^{2}A  (1/c^{2})∂^{2}A/∂t^{2}

∇(∇∙A +(1/c^{2})∂Φ/∂t )
= μ_{0}j.
(b) Lorentz condition:
∇∙A = (1/c^{2})∂Φ/∂t (We
pick an expression for ∇∙A.)
Simplification:
∇^{2}Φ  (1/c^{2})∂^{2}Φ/∂t^{2}
= ρ/ε_{0}, ∇^{2}A 
(1/c^{2})∂^{2}A/∂t^{2} = μ_{0}j.
are
now the differential equations satisfied by the potentials.
Φ, A_{x}, A_{y}, A_{z}
now satisfy the inhomogeneous wave equation.
(c) A and Φ are not unique.
A
> A + ∇ψ,
Φ > Φ  ∂ψ/∂t
is called a gauge transformation.
E and
B are invariant
under such a transformation.
Problem:
In the derivation of the wave
equations for A and Φ from Maxwell's equations in a vacuum, one gets at
one stage
∇^{2}A  (1/c^{2})∂^{2}A/∂t^{2}

∇(∇∙A +(1/c^{2})∂Φ/∂t )
= μ_{0}j.
and
∇^{2}Φ + ∂∇∙A/∂t
= ρ/ε_{0}.
This is made solvable by use of a gauge choice, using the fact that physics is
invariant under a gauge transformation. Write down the Lorentz gauge
condition and the resulting PDE's for Φ and A.
Solution:
 Concepts:
Maxwell's equations, scalar and
vector potential, the Lorentz condition, gauge transformations
 Reasoning:
We are asked to use the Lorentz condition to simplify the given PDE's.
 Details of the calculation:
Lorentz condition:
∇∙A = (1/c^{2})∂Φ/∂t
(we pick an expression for ∇∙A.)
Simplification:
∇^{2}Φ  (1/c^{2})∂^{2}Φ/∂t^{2}
= ρ/ε_{0}, ∇^{2}A 
(1/c^{2})∂^{2}A/∂t^{2} = μ_{0}j.
are now the differential equations satisfied by the potentials.
Φ, A_{x}, A_{y}, A_{z}
now satisfy the inhomogeneous wave equation.
A and Φ are not unique.
A > A + ∇ψ,
Φ > Φ 
∂ψ/∂t is called a
gauge transformation.
E and B are invariant under such a transformation.
Problem:
(a) Find the magnetic field associated with the potential
A(r,t)
= (b/2)n×r.
(b) Find the charge and current distributions that would lead to
A(r,t) = (bt/r^{3})r, Φ(r,t)
= 0.
(c) Determine the charge distribution that will give rise to the potential
Φ(r,t) = W_{0}exp(αr)/r.
Solution:
 Concepts:
Maxwell's equations, scalar and vector potential, Gauss' law
 Reasoning:
E = ∂A/∂t
 ∇Φ,
B = ∇×A, (SI units).
Given the potentials we are asked to find the fields.
 Details of the calculation:
(a) A_{x} = (b/2)(n_{y}z  n_{z}y), A_{y}
= (b/2)(n_{z}x  n_{x}z), A_{z} = (b/2)(n_{x}y
 n_{y}x).
B_{x} = ∂A_{z}/∂y
 ∂A_{y}/∂z =
(b/2)(n_{x} + n_{x}) = bn_{x}, B_{y} = bn_{y},
B_{z} = bn_{z}.
B = bn, a uniform field in the
n direction of strength b.
(b)
E = ∂A/∂t
 ∇Φ = (b/r^{3})r.
This is the field of a point charge at the origin with q
= 4πε_{0}b.
B =
∇×A = 0, ∇×B
= μ_{0}j + (1/c^{2})∂E/∂t
= 0, this implies that j = 0.
(c)
E(r) = ∇Φ(r) = W_{0}(α exp(αr)/r + exp(αr)/r^{2})(r/r).
E(r) is radial, we have a spherically symmetric charge
distribution.
∫_{surface of sphere of radius R}E∙ndA = E(R)4πR^{2}
= 4πW_{0}(1
+ αR)exp(αR) = Q_{inside}/ε_{0}.
As r > 0, Q_{inside} > 4πε_{0}W_{0},
there is a point charge at the origin.
For r ≠ 0, ρ(r) = ε_{0}∇^{2}Φ(r)
= ε_{0}∇^{2}(W_{0}exp(αr)/r)
= ε_{0}W_{0}(1/r^{2})(∂/∂r)[r^{2}(∂/∂r)(exp(αr)/r)].
ρ(r) = ε_{0}W_{0}α^{2}exp(αr)/r.
Therefore ρ(r) = ε_{0}W_{0}(4πδ(r)
 α^{2}exp(αr)/r).
Problem:
(a) State Maxwell's equations and prove that they are satisfied by E = ∂A/∂t,
B = ∇×A, provided
∇∙A = 0, ∇^{2}A
= (1/c^{2})∂^{2}A/∂t^{2}.
(b) Derive E and B when
A =
i a cos(k(z  ct))+ + j b sin(k(z  ct)).
(c) Verify that E and B are orthogonal and their directions rotate about
the zaxis with frequency kc/(2π).
Solution:
 Concepts:
Maxwell's equations,
E = ∂A/∂t
 ∇Φ,
B = ∇×A
 Reasoning:
Given ∇∙A = 0,
∇^{2}A  (1/c^{2})∂^{2}A/∂t^{2}
= 0, we are supposed to show that E = ∂A/∂t,
B = ∇×A satisfy Maxwell's equations.
 Details of the calculation:
(a) Maxwell's equations in SI units are
∇∙E = ρ/ε_{0},
∇×E = ∂B/∂t,
∇∙B = 0,
∇×B = μ_{0}j + (1/c^{2})∂E/∂t
Assume that E = ∂A/∂t,^{ }
B = ∇×A.
∇∙B = 0 automatically and
∇∙E = 0 from
∇∙A = 0, which implies that ρ = 0.
∇×E = ∂/∂t(∇×A)
= ∂B/∂t.
Only the
∇×B equation remains.
∇×B =
∇×(∇×A)
= ∇(∇∙A)
 ∇^{2}A =
∇^{2}A.
Given:
∇^{2}A  (1/c^{2})∂^{2}A/∂t^{2}
= 0
Therefore ∇×B
= (1/c^{2})∂^{2}A/∂t^{2}
= (1/c^{2})∂E/∂t,
which is the remaining equation. It implies that j = 0.
In free space with ρ = 0 and j = 0
Maxwell's equations are correctly obtained from a single vector potential
A
satisfying
∇∙A = 0,
∇^{2}A  (1/c^{2})∂^{2}A/∂t^{2}
= 0, Φ = 0.
(b) When A =
i a cos(k(z  ct)) +
j b sin(k(z
 ct)), we have
E = ∂A/∂t = i akc sin(k(z
 ct)) + j bkc cos(k(z  ct)).
B =
∇×A =
i kb cos(k(z  ct)) 
j
ak sin(k(z  ct)).
(c) E∙B = akc sin(k(z  ct)) kb cos(k(z  ct))  bkc
cos(k(z  ct)) ak sin(k(z  ct)) = 0.
E and
B are orthogonal.
At a given position along the zaxis, say z = 0,
E = i
akc sin(kct) + j bkc cos(kct) = i akc sin(ωt) +
j bkc cos(ωt).
The x and y coordinates oscillate 90^{o} out of phase with angular
frequency ω = kc, the tip of E moves clockwise on an ellipse.
B = i kb cos(ωt) +
j ak sin(ωt), the tip of
B
moves clockwise on an ellipse.
Problem:
(a) A point charge q rests at the origin.
A natural
choice of potentials for this static problem is V(r,t) = q/(4πε_{0}r),
A(r,t) = 0.
Consider the gauge transformation with λ(r,t) = qt/(4πε_{0}r)
+ k/r^{3},^{
}where k is a constant.
Calculate the transformed potentials and fields. Discuss
the result.
(b) Find E and B given the potentials
A(ρ,φ,z) = k C ln(ρ^{2})/(4π), V = 0.
Which charge or current
distributions produce these fields? Can you find a gauge transformation which
leaves the transformed vector potential A(ρ,φ,z) = 0?
Solution:
 Concepts:
Gauge transformations
 Reasoning:
A and Φ are
not unique.
A > A + ∇λ,
V > V  ∂λ/∂t is called a gauge
transformation.
 Details of the calculation:
(a) Transformed potentials:
A_{r}(r,t) = ∂λ/∂r
= qt/(4πε_{0}r^{2})
 3k/r^{4}. The other components of A are zero.
V(r,t) = q/(4πε_{0}r)
 q/(4πε_{0}r)
= 0.
E = ∂A/∂t
 ∇V, B =
∇×A,
(SI units).
Transformed fields:
E_{r} = ∂A_{r}/∂t = q/(4πε_{0}r^{2}),
the other components of E are zero.
B = ∇×A
= 0 since A is irrotational. It has only a radial component which does
not depend on θ and φ.
E and B are invariant under a gauge
transformation.
(b) B
=
∇×A = ∂A_{z}/∂ρ (φ/φ) = C/(2πρ) (φ/φ),
E = 0.
This is the magnetic field produced by a current I = C/μ_{0} in an
infinitely long, straight wire. No free charges are present.
A gauge transformation leaving
A' = 0 does not
exist since B = ∇×A
is not equal to zero.
Problem:
(a) Write down expressions for the electric field E and the
magnetic field B in terms of the scalar potential Φ
and the vector potential A.
(b) Given Φ = 0 and A =
k [μ_{0}α/(4c)](ct
x)^{2} for x < ct, A = 0 for x > ct, find
E and
B.
(c) Write down the boundary conditions for the normal component of
E and
the tangential component of B at the x = 0 interface.
(d) Find the surface charge density and the surface current density at the
x = 0 interface. (Assume media with ε_{0}
and μ_{0} on both sides of the interface.)
Solution:
 Concepts:
Maxwell's equations, scalar and vector potential
 Reasoning:
We are supposed to write down E and
B in terms of the scalar
and vector potential.
 Details of the calculation:
(a)
E = ∂A/∂t
 ∇Φ,
B = ∇×A
(b)
E = k [μ_{0}α/2](ct
x), B = ±j [μ_{0}α/(2c)](ct
x), It x > 0 B points into the ydirection, if x < 0
B points into the ydirection
(c and d)
(E_{2}  E_{1})∙n_{2} = σ/ε_{0},,
E
is continuous at x = 0, there is no surface charge density.
(B_{2}  B_{1})∙t =
μ_{0}k∙n, B is discontinuous at x =
0, k_{surface} = αt
k.
Problem:
Show that in
free space with ρ = 0 and j = 0
Maxwell's equations are correctly obtained from a single vector potential
A
satisfying
∇∙A = 0,
∇^{2}A  (1/c^{2})∂^{2}A/∂t^{2}
= 0, Φ = 0.
Solution:
 Concepts:
Maxwell's equations,
E = ∂A/∂t
 ∇Φ,
B = ∇×A
 Reasoning:
Given ∇∙A = 0,
∇^{2}A  (1/c^{2})∂^{2}A/∂t^{2}
= 0, Φ = 0, we are supposed to obtain Maxwell's
equations.
 Details of the calculation:
Φ = 0, therefore E = ∂A/∂t,^{
}
B = ∇×A.
∇∙B = 0 automatically and
∇∙E = 0 from
∇∙A = 0.
∇×E = ∂/∂t(∇×A)
= ∂B/∂t.
Only the ∇×B equation remains.
∇×B =
∇×(∇×A)
= ∇(∇∙A)
 ∇^{2}A =
∇^{2}A.
Given:
∇^{2}A  (1/c^{2})∂^{2}A/∂t^{2}
= 0.
Therefore ∇×B
= (1/c^{2})∂^{2}A/∂t^{2}
= (1/c^{2})∂E/∂t,
which is the remaining equation.
Problem:
(a) Consider the following vector potential A:
A_{x} = 0, A_{y} = bx, A_{z} = 0.
Find the corresponding magnetic field.
(b) Consider the following vector potential A':
A'_{x} = by, A'_{y} = 0, A'_{z} = 0.
Find the corresponding magnetic field.
(c) A' = A +
∇f(x, y,z). Find f.
(d) The Lagrangian of a particle with mass m and charge q moving in the
presence of a magnetic field is
L = ½ mv^{2} + qv∙A.
Using the vector potential from part (a) or part (b), write down Lagrange's
equations of motion, the Hamiltonian, and Hamilton's equations of motion.
(e) Solve the equations of motion.
Solution:
 Concepts:
The vector potential
 Reasoning:
The vector potential is not unique. It enters into the Lagrangian, but
the equations of motion do not depend on the gauge.
 Details of the calculation:
(a) B = ∇×A = b
k.
(b) B = ∇×A' = b
k.
(c) ∂f/∂x = by, ∂f/∂y = bx, ∂f/∂z = 0.
f = byx + constant.
(d) Using the vector potential from part (a):
L = ½ m(v_{x}^{2} + v_{y}^{2} + v_{z}^{2})
+ qv_{y}bx.
∂L/∂v_{x} = mv_{x} = p_{x} ∂L/∂v_{y} =
mv_{v} + qbx = p_{y} = constant, ∂L/∂v_{z} =
mv_{z} = p_{z} = constant.
Lagrange's equations of motion:
p_{z} = mv_{z} = constant, dv_{z} /dt = a_{z}
= 0.
p_{y} = mv_{v} + qbx = constant, dv_{y} /dt = a_{y}
= qbv_{x}/m.
d/dt(∂L/∂v_{x})  ∂L/∂x = 0, a_{x} = qbv_{y}/m.
H(x,p) = p_{x}v_{x} + p_{y}v_{y} +
p_{z}v_{z}  L
= p_{x}^{2}/m + p_{y}(p_{y}  qbx)/m + p_{z}^{2}/m  ½p_{x}^{2}/m  ½(p_{y}
qbx)^{2}/m  ½p_{z}^{2}/m  qv_{y}bx
= ½p_{x}^{2}/m + ½(p_{y}  qbx)^{2}/m + ½p_{z}^{2}/m.
Note: p is the canonical, not the mechanical momentum.
Hamilton's equations of motion: ∂H/∂p_{x} = dx/dt, ∂H/∂x
= dp_{x}/dt, similar for the other components.
v_{x} = p_{x}/m, v_{y} = (p_{y}  qbx)/m,
v_{z} = p_{z}/m,
dp_{x}/dt = qb(p_{y}  qbx)/m, dp_{y}/dt = 0, dp_{z}/dt = 0.
This yields a_{z} = 0, a_{y} = qbv_{x}, a_{x}
= qb(p_{y}  qbx)/m^{2} = qbv_{y}/m.
(e) a_{z} = 0, v_{z} = v_{z0}, z = z_{0}
+ v_{z0}t.
a_{y} = qbv_{x}, a_{x} = qbv_{y}/m.
x = Acos(ωt + φ) + x_{0}, y = Asin(ωt + φ) + y_{0}, v_{x}
= ωAsin(ωt + φ), v_{y} = ωAcos(ωt + φ),
a_{x} = ω^{2}Acos(ωt + φ) = ωv_{y}, a_{y}
= ω^{2}Asin(ωt + φ) = ωv_{x}, ω = qb/m.
The initial conditions determine x_{0}, y_{0}, A and φ.